Describe the differences between the graph of y equals negative 3 over 4 open parentheses x minus 6 close parentheses squared plus 4 and the standard position graph of y = x2.
please help

Question

Describe the differences between the graph of y equals negative 3 over 4 open parentheses x minus 6 close parentheses squared plus 4 and the standard position graph of y = x2.
please help

ranga · Answer

$$y = -\frac{ 3 }{ 4}(x-6)^2 + 4 \quad \text{compared with} \quad y = x^2$$

ranga · Answer

Replacing x in the equation y = x^2 with (x-6) is equivalent to shifting the graph y = x^2 to the RIGHT by 6 units.

Changing y = x^2 to y = x^2 + 4 is equivalent to shifting the graph UP by 4 units.

continued...

ranga · Answer

In y = x^2, the coefficient of x^2 is positive.  Therefore, the graph, which is a parabola, opens upwards.  If the coefficient is negative, the graph will be an inverted parabola which will open downward.

So y = -x^2 is a parabola that opens downward.  
If the -1 coefficient were changed to a fraction, such as -3/4, the graph will be compressed vertically (a little bit squished).

ranga · Answer

So starting with y = x^2 you can get the other graph by shifting it to the right by 6 units, lifting it up by 4 units, flipping it so it opens downward and squishing it slightly.
y = x^2 has vertex at (0,0).  It has a minimum at the vertex and the minimum value is 0.  It is a parabola that opens upward.  It touches the x-axis at just one point.
y = -3/4(x-6)^2 + 4  has the vertex at (6,4).  Because it is an inverted parabola, it attains its maximum at x = 6 and the maximum value is 4.  It cuts the x axis at two points.